Suppose `alpha, beta` are roots of `ax^(2)+bx+c=0` and `gamma, delta` are roots of `Ax^(2)+Bx+C=0`.
If `alpha,beta,gamma,delta` are in GP, then common ratio of GP is

Options

A. `sqrt(((bA)/(aB)))`
B. `sqrt(((aB)/(bA)))`
C. `sqrt(((bC)/(cB)))`
D. `sqrt(((cB)/(bC)))`

Explanation

To solve the problem, we need to find the common ratio \( r \) of the geometric progression (GP) formed by the roots \( \alpha, \beta, \gamma, \delta \) of the given quadratic equations.

### Step-by-Step Solution:

1. **Identify the Roots**:
- Let \( \alpha \) and \( \beta \) be the roots of the equation \( ax^2 + bx + c = 0 \).
- Let \( \gamma \) and \( \delta \) be the roots of the equation \( Ax^2 + Bx + C = 0 \).

2. **Roots in GP**:
- Since \( \alpha, \beta, \gamma, \delta \) are in GP, we can express them in terms of \( \alpha \) and the common ratio \( r \):
- \( \beta = \alpha r \)
- \( \gamma = \alpha r^2 \)
- \( \delta = \alpha r^3 \)

3. **Sum of Roots**:
- For the first quadratic \( ax^2 + bx + c = 0 \):
- The sum of the roots \( \alpha + \beta = -\frac{b}{a} \).
- Substituting \( \beta \):
\[ \alpha + \alpha r = -\frac{b}{a} \]
\[ \alpha(1 + r) = -\frac{b}{a} \quad \text{(Equation 1)} \]

- For the second quadratic \( Ax^2 + Bx + C = 0 \):
- The sum of the roots \( \gamma + \delta = -\frac{B}{A} \).
- Substituting \( \gamma \) and \( \delta \):
\[ \alpha r^2 + \alpha r^3 = -\frac{B}{A} \]
\[ \alpha r^2(1 + r) = -\frac{B}{A} \quad \text{(Equation 2)} \]

4. **Divide the Two Equations**:
- Divide Equation 2 by Equation 1:
\[ \frac{\alpha r^2(1 + r)}{\alpha(1 + r)} = \frac{-\frac{B}{A}}{-\frac{b}{a}} \]
- Simplifying gives:
\[ r^2 = \frac{B}{A} \cdot \frac{a}{b} \]

5. **Finding the Common Ratio**:
- Taking the square root of both sides, we find the common ratio \( r \):
\[ r = \sqrt{\frac{B \cdot a}{b \cdot A}} \]

### Final Answer:
The common ratio \( r \) of the geometric progression is:
\[ r = \sqrt{\frac{B \cdot a}{b \cdot A}} \]